Question

Hyperbola question.... did I do this right?
(question below)

Answer 1

25. Find the equation of the hyperbola.
Vertices at (0, -6) and (0, 6); asymptote the line y = 2x
I know that a = 6
I used y-k = \(\pm \dfrac ba\)(x-h) to find b
I know the center is (0, 0) because y - k = y and x- h = x
y = \(\pm \dfrac ba\)(x)

y = \(\pm \dfrac b6\)(x)
From that I got that b = 12
Then used \(b^2 = c^2 - a^2\) to find \(c^2\)
144 = \(c^2\) - 36
180 = \(c^2\)

Then used \(\dfrac{y^2}{a^2} - \dfrac{x^2}{b^2}\) = 1 to get the equation

Equation: \(\dfrac{y^2}{36} - \dfrac{x^2}{180} = 1\)

Answer 2

doesn't look correct

Answer 3

compare the given asymptote with
\[y = \pm \dfrac{a}{b}x\]

Answer 4

\(\dfrac{a}{b}=2 \) and \(a=6\) give you \(b = 3\)

Answer 5

With \(a = 6, b = 3\), the eqn should be
\[\dfrac{y^2}{36}-\dfrac{x^2}{9}=1\]

Answer 6

http://www.wolframalpha.com/input/?i=asymptotes+y^2%2F36+-+x^2%2F9+%3D+1

Answer 7

ah, I see where I went wrong, thanks :)